I need help solving the question below:
$$ 2x^ \frac{1}{4} = \frac {64} {x} $$
I know the answer is 16 but I'm not sure how to get to it. Can you explain how to get the answer so I can solve similar questions
I need help solving the question below:
$$ 2x^ \frac{1}{4} = \frac {64} {x} $$
I know the answer is 16 but I'm not sure how to get to it. Can you explain how to get the answer so I can solve similar questions
$$2x^{\dfrac14}\cdot x=32\implies x^{\left(\dfrac14+1\right)}=32=2^5\implies x^{\dfrac54}=2^5$$
$$\implies x^{\dfrac14}=(2^5)^{\dfrac15}$$
Now one of the five values of $\displaystyle(2^5)^{\dfrac15}$ is $2$
In that case $\displaystyle x^{\dfrac14}=2\implies (x^{\dfrac14})^4=2^4\implies x=2^4$
$2x^{\frac{1}{4}}(x) = 64 \Rightarrow 2x^\frac{1}{4}x^\frac{4}{4} = 64 \Rightarrow 2x^\frac{5}{4} = 64 \Rightarrow x^{\frac{5}{4}}=32 \Rightarrow x =(32)^{\frac{4}{5}}$
As, $2^5=32 \Rightarrow x = (2^5)^{\frac{4}{5}} = 2^4 = 16$
Multiply both sides by $x$, and you get $$2x^{\frac{5}{4}}=64$$ and then $$x^{\frac{5}{4}}=32$$ $$x^{\frac{5}{4}}=2^5$$ $$x^{\frac{1}{4}}=2$$ $$x=2^4=16$$
$x$ is positive, so you can take the logarithm and get $\ln2+\frac14\ln x=\ln(64)-\ln x$ and solve for $\ln x$. Since $\ln(64)=6\ln 2$, you get easily to $\ln x=4\ln 2$. $x=16$ follows after taking the exponential.